Contoh Soal 10 Turunan Fungsi Trigonometri (Bagian 3)

$\begin{array}{l}\\ 46.& \text{Turunan kedua dari}f(x)=x^3-\sin 3x\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 6x^2+9\sin 3x\\ \text{b}.& 3x^2+6\sin 3x\\ \text{c}.& 3x-9\sin 3x\\ \text{d}.& 6x+9\sin 3x\\ \text{e}.& 9x-6\sin 3x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}f(x)& =x^3-\sin 3x\\ f^{\prime }(x)& =3x^2-3\cos 3x\\ f^{″}(x)& =6x+9\sin 3x\end{array}\end{array}$

$\begin{array}{l}\\ 47.& \text{Diketahui fungsi}g(x)={\displaystyle \frac{1-\cos x}{\sin x}.\text{Nilai}}\\ & \text{turunan kedua saat}x={\displaystyle \frac{\pi }{4}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& \sqrt{2}+4\\ \text{b}.& 2\sqrt{2}-3\\ \text{c}.& 2\sqrt{2}+3\\ \text{d}.& 3\sqrt{2}-4\\ \text{e}.& 3\sqrt{2}+4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}g(x)& ={\displaystyle \frac{1-\cos x}{\sin x}}\\ g^{\prime }(x)& ={\displaystyle \frac{\sin x(\sin x)-\cos x(1-\cos x)}{{\sin }^{2}x}}\\ & ={\displaystyle \frac{{\sin }^{2}x-\cos x+{\cos }^{2}x}{{\sin }^{2}x}}\\ & ={\displaystyle \frac{1-\cos x}{{\sin }^{2}x}}\\ g^{″}(x)& ={\displaystyle \frac{\sin x({\sin }^{2}x)-2\sin x\cos x(1-\cos x)}{{\sin }^{4}x}}\\ & ={\displaystyle \frac{\sin x({\sin }^{2}x)-\sin 2x(1-\cos x)}{{\sin }^{4}x}}\\ & ={\displaystyle \frac{\sin {\displaystyle \frac{\pi }{4}({\sin }^2{\displaystyle \frac{\pi }{4})-\sin 2{\displaystyle \frac{\pi }{4}(1-\cos {\displaystyle \frac{\pi }{4})}}}}}{{\sin }^4{\displaystyle \frac{\pi }{4}}}}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{1}{\sqrt{2}}}\right){\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}^2-1.(1-\left({\displaystyle \frac{1}{\sqrt{2}}}\right))}{{\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}^4}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{2}{\displaystyle \frac{1}{\sqrt{2}}-1+{\displaystyle \frac{1}{\sqrt{2}}}}}}{{\displaystyle \frac{1}{4}}}\times {\displaystyle \frac{4}{4}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{2}{\sqrt{2}}-4+\frac{4}{\sqrt{2}}}}{1}}\\ & ={\displaystyle \frac{6}{\sqrt{2}}-4=3\sqrt{2}-4}\end{array}\end{array}$

$\begin{array}{l}\\ 48.& \text{Turunan kedua fungsi}f(x)={\sin }^{2}x-{\cos }^{2}x\\ & \text{adalah}{f}^{″}(x)=....\\ & \begin{array}{l}\\ \text{a}.& 6\sin 2x\\ \text{b}.& 4\cos 2x\\ \text{c}.& 2\cos 2x\\ \text{d}.& -2\cos 2x\\ \text{e}.& -4\cos 2x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}f(x)& ={\sin }^{2}x-{\cos }^{2}x\\ f^{\prime }(x)& =2\sin x\cos x-2\cos x(-\sin x)\\ & =2\sin x\cos x+2\sin x\cos x\\ & =2(2\sin x\cos x)=2\sin 2x\\ f^{″}(x)& =2.2\cos 2x=4\cos 2x\end{array}\end{array}$

$\begin{array}{l}\\ 49.& \text{Diketahui}f(x)=\sqrt{\sin x}.\text{Jika}{f}^{″}(x)\\ & \text{adalah turunan keduafungsi}f,\text{maka}\\ & \text{nilai dari}{f}^{″}\left({\displaystyle \frac{\pi }{2}}\right)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{1}{2}}\\ \text{b}.& -{\displaystyle \frac{1}{4}}\\ \text{c}.& 0\\ \text{d}.& {\displaystyle \frac{1}{4}}\\ \text{e}.& {\displaystyle \frac{1}{2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}f(x)& =\sqrt{\sin x}={\sin }^{\frac{1}{2}}x\\ f^{\prime }(x)& ={\displaystyle \frac{1}{2}{\sin }^{-\frac{1}{2}}x.\cos x={\displaystyle \frac{\cos x}{2{\sin }^{\frac{1}{2}}x}}}\\ f^{″}(x)& ={\displaystyle \frac{-\sin x(2{\sin }^{\frac{1}{2}}x)-\cos x\left(2.{\displaystyle \frac{1}{2}{\sin }^{-\frac{1}{2}}x.\cos x}\right)}{4\sin x}}\\ & ={\displaystyle \frac{-2\sin x\sqrt{\sin x}-{\displaystyle \frac{{\cos }^{2}x}{\sqrt{\sin x}}}}{4\sin x}}\\ f^{″}\left({\displaystyle \frac{\pi }{2}}\right)& ={\displaystyle \frac{-2\sin {\displaystyle \frac{\pi }{2}.\sqrt{\sin {\displaystyle \frac{\pi }{2}}}-{\displaystyle \frac{{\cos }^2{\displaystyle \frac{\pi }{2}}}{\sin {\displaystyle \frac{\pi }{2}}}}}}{4\sin {\displaystyle \frac{\pi }{2}}}}\\ & ={\displaystyle \frac{-2.1.1-0}{4.1}=-\frac{1}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 50.& \text{Jika}f(x)={\tan }^2(3x-2)\text{maka}{f}^{″}(x)=....\\ & \begin{array}{l}\\ \text{a}.& 36{\tan }^2(3x-2){\sec }^2(3x-2)\\ & -18{\sec }^4(3x-2)\\ \text{b}.& 36{\tan }^2(3x-2){\sec }^2(3x-2)\\ & +18{\sec }^2(3x-2)\\ \text{c}.& 36{\tan }^2(3x-2){\sec }^2(3x-2)\\ & +18{\sec }^4(3x-2)\\ \text{d}.& 18{\tan }^2(3x-2){\sec }^2(3x-2)\\ & +36{\sec }^4(3x-2)\\ \text{e}.& 18{\tan }^2(3x-2){\sec }^2(3x-2)\\ & +18{\sec }^4(3x-2)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}f(x)& ={\tan }^2(3x-2)\\ f^{\prime }(x)& =2\tan (3x-2){\sec }^2(3x-2)(3)\\ & =6\tan (3x-2){\sec }^2(3x-2)\\ f^{″}(x)& =6{\sec }^2(3x-2).(3){\sec }^2(3x-2)\\ & +6\tan (3x-2).2\sec (3x-2).\sec (3x-2)\tan (3x-2)(3)\\ & =18{\sec }^4(3x-2)\\ & +36{\tan }^2(3x-2){\sec }^2(3x-2)\end{array}\end{array}$